Compute the integral of $ z^2+ z z*$ where $C=\{ z | |z|=1, 0<argz<\pi \}$

Question

(Z* is z-conjugate)

This has been my attempt: I have let $z=e^{(it)}$ for $0<t<\pi$

So then i get that $z^2+ z z* = e^{(2it)}+1$ and that $ dz=ie^{(it)} dt$ I formed my integral with these facts, to attain $ie^{(3it)}+ie^{(it)}$ from 0 to $\pi$ Calculating this as a normal integral got me -8/3 However I do not feel I have gone about this the right way??

Answer 1

Your computation is fine.

Here is another way: Along $C$ one has $z\bar z=1$, hence $$\int_C(z^2+z\bar z)\>dz =\int_C(z^2+1)\>dz=\left({z^3\over3}+z\right)\Biggr|_1^{-1}=-{8\over3}\ .$$